Write the Lewis structure for each molecule or ion. a. H 3 COH d. O 2 2-

Step 1: Identify the total number of valence electrons for each molecule or ion. For H 3 COH, count the valence electrons from H (1 each), C (4), and O (6). For O 2 2- , count the valence electrons from each O (6 each) and add 2 extra electrons for the 2- charge.. Step 2: Arrange the atoms to show how they are connected. For H 3 COH, place C in the center with O and H atoms around it. For O 2 2- , place the two O atoms next to each other.. Step 3: Connect the atoms with single bonds initially. For H 3 COH, connect C to O and C to each H. For O 2 2- , connect the two O atoms with a single bond.. Step 4: Distribute the remaining valence electrons to satisfy the octet rule (or duet for hydrogen). Start by completing the octets for the outer atoms first, then the central atom. Adjust bonds if necessary to ensure all atoms have a complete octet.. Step 5: Check the formal charges on each atom to ensure the most stable structure. Adjust the structure if needed to minimize formal charges, ensuring the overall charge of the ion is correct.

Write the Lewis structure for each molecule or ion. a. H 3 COH d. O 2 2-

hey everyone, we're asked to draw the lewis structures of the following molecules and or ions. So first we're going to want to calculate the number of valence electrons will have in total. Starting off with a We know that iodine is in our group 70. So this will give us seven valence electrons since we only have one of them For flooring. Since we have to we're going to multiply two times seven since it's also in our group seven a. Which will get us to 14 Adding these two. Up we get a total of 21 valence electrons. But since we have that negative one charge, we're going to have to add one valence electron. So we should get 22 when we draw this out, iodine is going to be our central atom since it's less electro negative than our flooring and we're going to complete the octave of our flooring first before adding on the lone pairs onto our iodine. So reaching all the way up to 22 valence electrons, we're going to end up With three lone pairs surrounding our iodine. And since we have that negative one charge, we can't forget to add the brackets and that negative one charge on the side for B. We know that xenon is in our group ate a so we're going to have eight valence electrons and for browning, we know it's in our group 78. So seven times four is going to come up to 28 Adding these two up, we're going to get £36 electrons. Now when we draw this out, Zenon is going to be our central atom And we're going to surround it with four row means And the same way we did in our first ion we're going to complete bro means octavius first before adding on the lone pairs onto our xenon. So to reach 36, we're going to end up with two lone pairs on Xenon. Moving on to see We know that nitrogen is going to contribute five valence electrons since it's in our group five a. And oxygen is going to contribute six, Which will get us to a total of 11. And since we have that odd number, this is basically telling us that we're going to have a radical electron. So we have our nitrogen bonded to our oxygen And we can create a double bond between the two And adding a two lone pairs onto our oxygen. We can also add one lone pair on our nitrogen and to complete the 11 valence electrons, we're going to have to add one radical electron on top of our nitrogen. So when we calculate the valence electrons surrounding each atom, we have 12345. Which makes sense since nitrogen contributed vibe. And for oxygen 1, 2, 3, 4, 5 and six. So this also makes sense. Moving on to D. We know that boron is going to contribute three since it's in our group three A and hydrogen is also going to contribute three since one times three equals three. So this will get us to a total of six. When we draw this out, we're going to have boron as our central atom since hydrogen can't be our central atom and we're going to end up with the following structure. So we have an incomplete octet here. But this is an exception to the rule. So this is okay and lastly moving on to eat, we know that we have oxygen and oxygen is going to contribute six valence electrons. Phosphorus is going to contribute five since it's in our group 58 and we have three of chlorine which will contribute Adding these values up. We end up with a structure that has 32 valence electrons. Now phosphorus is going to be our central atom And it's going to be surrounded by three chlorine and one oxygen. And it's going to be double bonded to our oxygen, since we don't have a formal charge on our oxygen. So that means we'll have two lone pairs on our oxygen and three lone pairs on two, all of our chlorine, so we can obey our octet rule. And once we do all that We end up with 32 valence electrons and phosphorus is going to actually have 10 electrons surrounding it. And this is okay because this is an expanded octet and this is also an exception to the rule. So I hope this made sense and let us know if you have any questions

Ch.9 - Chemical Bonding I: The Lewis Model

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Ch.9 - Chemical Bonding I: The Lewis Model

Problem 6a,d

Chapter 9, Problem 6a,d

Write the Lewis structure for each molecule or ion. a. H₃COH d. O₂^2-

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Step 1: Identify the total number of valence electrons for each molecule or ion. For H3COH, count the valence electrons from H (1 each), C (4), and O (6). For O22-, count the valence electrons from each O (6 each) and add 2 extra electrons for the 2- charge.

Step 2: Arrange the atoms to show how they are connected. For H3COH, place C in the center with O and H atoms around it. For O22-, place the two O atoms next to each other.

Step 3: Connect the atoms with single bonds initially. For H3COH, connect C to O and C to each H. For O22-, connect the two O atoms with a single bond.

Step 4: Distribute the remaining valence electrons to satisfy the octet rule (or duet for hydrogen). Start by completing the octets for the outer atoms first, then the central atom. Adjust bonds if necessary to ensure all atoms have a complete octet.

Step 5: Check the formal charges on each atom to ensure the most stable structure. Adjust the structure if needed to minimize formal charges, ensuring the overall charge of the ion is correct.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They use dots to represent valence electrons and lines to represent bonds between atoms. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting the behavior of molecules in chemical reactions.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in determining how atoms bond with each other. The number of valence electrons influences the atom's reactivity and the types of bonds it can form. In Lewis structures, counting valence electrons helps ensure that the correct number of electrons is represented in the diagram, which is vital for accurate molecular representation.

Formal Charge

Formal charge is a concept used to determine the distribution of electrons in a molecule and to assess the stability of a Lewis structure. It is calculated by comparing the number of valence electrons in an isolated atom to the number of electrons assigned to it in the Lewis structure. A lower formal charge on atoms generally indicates a more stable structure, guiding chemists in choosing the most favorable Lewis structure for a given molecule.

Write the Lewis structure for each molecule or ion. a. H3COH d. O22-